3 Answers
3

You are doing fine up to $a^2-2a+b^2 \ge 0$. Then complete the square to get $(a-1)^2+b^2 \ge 1$. These are the points on or outside the circle with centre $(1,0)$, radius $1$.

For the problem about $\Im\left(\frac{z+i}{2i}\right)\le 0$, one way to start is to let $z=a+bi$ and compute. We have $z+i=a+(b+1)i$. Divide by $2i$. We get $\frac{b+1}{2}-\frac{a}{2}i$. The imaginary part of this is $\le 0$ precisely if $a\ge 0$. We needn't have bothered adding the $i$, since we are dividing by $2i$, so the imaginary part will not be affected. And we needn't have bothered with $b$ either.

For the second one: first of all, notice that the expression $Im(\frac{z+i}{2i})$ is just $Im(\frac{z}{2i}+\frac{1}{2})$, and the latter part - being purely real - doesn't contribute anything to the imaginary part of the expression, so you're really just looking at the set $Im(\frac{z}{2i})\leq 0$; since $1/i = -i$, this is the same as $Im(-iz/2)\leq 0$. But you can multiply by the real constant -2 (since $Im(r\cdot x)=r\cdot Im(x)$ for any real $r$) to get $Im(iz)\geq 0$. Now, $Im(iz) = Re(z)$, so this is just $Re(z)\geq 0$.

The other way is just to plough through the algebra. Suppose $z=a+bi$, with $a$ and $b$ real; then $z+i$=$a+(b+1)i$, $(z+i)/2i$ = $(b+1)/2-ai/2$, $Im(z+i)/2i=-a/2$, and $\left(Im(z+i)/2i \leq 0\right) \equiv \left(-a/2\leq 0\right)\equiv \left(a\geq 0\right)$.

For the second one, $\text{Im} \left( \dfrac{x+iy + i}{2i} \right) = \text{Im} \left( \dfrac{y+1}{2} -i \dfrac{x}{2} \right) = -\dfrac{x}{2} \leq 0$. This means that $x \geq 0$. Hence, the region is the half plan to the right of the imaginary axis i.e. complex of numbers whose real part is non-negative.